Dear Professor Cram:

A man deposits 1200 at the end of each year into a savings account that earn 6.75% interest compounded annually. After 4 years he made no more deposits. What would be the balance in the account 3 years from the last deposit?

Eva, Faulkner University

Thanks for your question, Eva.

This is a future value of an annuity problem, complicated by making payments at the END of the periods, but not payments for the whole time. So we have to break it up into pieces: years with deposits, and years without deposits. First, let's lay out what we know about the payments. I usually do this on a timeline, but those are hard to type….

- End of first year – deposit $1,200
- End of second year – deposit $1,200
- End of third year – deposit $1,200

"After 4 years he made no more deposits." Hmmm – so did he make the 4th deposit or didn't he? I hate problems that are ambiguous. If you had his bank statement, it wouldn't be ambiguous… OK, I am going to assume he DID make the 4th deposit at the end of the 4th year, and after 4 years (of deposits) he made no more deposits.

- End of fourth year – deposit $1,200

The next thing we know is we flash forward three years from the last deposit (hmmm, did he really make that 4th deposit? makes a difference in the answer…) and we need to know the balance at the end of the 7th year.

Here is one way to solve this problem: Let's build it a year at a time.

At the end of the first year – he just has the deposit, no interest yet. OK, that is $1,200.

At the end of the second year, he should have the first deposit, plus a year's worth of interest at 6.75% ($1,200.00 x 0.065 = $78.00), plus the new deposit of $1,200.00. Let's add those up:

- $1,200.00 1st deposit

$1,200.00 new deposit at the end of the second year

__+ $ 78.00__interest during the second year

$2,478.00 at the end of the second year

This third year, he will be due interest on that whole balance of $2,478.00 from last year:

- ($2,428.00 x 6.75% = 161.07) [it went up by MORE than the 78.00 per $1,200 because of interest on interest – now we have compounding going on!]

- $2,478.00 at the end of the second year – our starting balance

$1,200.00 deposit at the end of the third year

__+ $161.07__interest during the third year

$3,839.07 at the end of the third year

You get the idea. Now repeat for fourth year (if he really made that 4th deposit – darned ambiguous word problems).

Once you have the ending balance after the last deposit, now we just add interest each year for three more years. You can either:

1. Take the ending balance from those payment years and now add 6.75% to it for each year and repeat three times; OR

2. If you can handle exponents, you can do all three years in one fell swoop. For each year it is the old balance x 1.0675 for the new ending balance. For three years that is the old balance x 1.0675 x 1.0675 x 1.0675 which is the old balance x (1.0675)^{3}.

I hope this helps. Let us know if you need anything else.

Good Studying,

Professor Cram